Question

A ball is projected horizontally with a velocity of $4m{s}^{-1}$ form the top of a tower. The velocity of the ball after $0.7s$ is (Taking $g=10m/s^{(}-2)$)

• A. $1m{s}^{-1}$
• B. $10m{s}^{-1}$
• C. $8m{s}^{-1}$
• D. $3m{s}^{-1}$

Answer 1

The correct option is C $8m{s}^{-1}$

using $v=u+at$ in both x and y direction

$a_x=0,a_y=g=10$

$v_x=u_x=4$

$v_y=u_y+10t$

$v_y=7$

$v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{4^2+7^2}$

$v=8.06$